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Classify all rings (with unit or not) whose additive group is cyclic.

Let $g$ be the generator of the additive group. Then we have a surjective group homomorphism $\phi$ from $\mathbb Z$ to $A$ which sends $1$ in $g$. So $A$ is isomorphic, as group, to some $\mathbb Z/n\mathbb Z$, $n\in \mathbb Z$ with isomorphism $\bar\phi$. One also gets (by applying the distributive property) that $\phi(n)\phi(m)=\phi(nm) g^2$.

What can be said now? Non-isomorphic rings are those where the additive order of $g$ is different. But also if that order is the same, I think there can be non-isomorphic rings. My claim is that the multiplicative order (in the ring $\mathbb Z/n\mathbb Z$) of $\bar\phi^{-1}(g^2)$ classifies these rings. Is this true?

And if not, can someone give me a clue on how to deal with this very interesting problem?

user26857
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W.Rether
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1 Answers1

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If you read Hungarian, then you can find this exercise as Problem 32 (for the finite case) and Problem 33 (for the infinite case) on page 79 of the 5th edition of

Absztrakt algebrai feladatok (Exercises in Abstract Algebra), Polygon, Szeged, 2005, by Maria Balintne-Szendrei, Gabor Czedli and Agnes Szendrei.

The solution can be found on page 405.

In any case, the answer is that any $n$-element ring with cyclic additive group is isomorphic to $d\mathbb Z_{dn}$ for some divisor $d$ of $n$, and different divisors yield nonisomorphic rings. (When $d=n$ you get a zero ring.) An infinite ring with cyclic additive group is a zero ring or is isomorphic to $d\mathbb Z$ for some positive $d$, and different choices of $d$ yield nonisomorphic rings.

Keith Kearnes
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  • So, in particular, for unital rings, there is only one isomorphism class for each $n$, namely the usual $\mathbb{Z}/n\mathbb{Z}$, right? – Anakhand Jan 26 '24 at 12:32